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How many grams of KI are in 25.0ml of a 3.0 (mv) KI solution?
55 ml of a 4.05 M solution of KI solution contains 55*4.05=222.75 millimoles. 20.5 ml of the diluted solution contains 3.8g of KI,so no.of moles of KI=3.8/(mol.wt of KI=165.9) is 22.9 millimoles. molarity of final diluted solution=22.9/20.5=1.117M since the no. of moles of KI present in initial and final solution are same. let.V(in ml) be the final volume of diluted solution. 222.75/V=1.117 V=199.41 ml final volume =199.41 ml
What mass of KI is present in 276 mL of a 0.50 M solution?
V*c*M = 276(mL)*0.001(L/mL)*0.50(mol KI/L)*166.0(g KI/mol KI) = 22.9 g KI
What is the molarity of 67.94 mL aqueous solution containing 2.822 g of KI?
Molarity = moles of solute/Liters of solution. get moles KI 2.822 grams KI (1 mole KI/166 grams) = 0.017 moles KI ( 67.94 ml = 0.06794 Liters ) Molarity = 0.017 moles KI/0.06794 Liters = 0.2502 M KI
To 225 mL of a 0.80M solution of KI a student adds enough water to make 1.0 L of a more dilute KI solution What is the molarity of the new solution?
0.18M
What is the molarity of a solution prepared by dissolving 2.41 g of potassium iodide KI in 100 mL of water?
Find moles potassium iodide first.2.41 grams KI (1 mole KI/166 grams) = 0.01452 moles KIMolarity = moles of solute/Liters of solution ( 100 ml = 0.1 Liters )Molarity = 0.01452 moles KI/0.1 Liters= 0.145 M KI solution================
How many moles of KI are present in 500 ml of a 0.2M solution?
The answer is 0,1 mol.
HOW TO prepare 0.1 n iodide solution in 500 ml water?
0.1 N iodide would be 0.1 moles of the iodide salt (e.g. KI) per liter of solution. For 500 ml, you would need 0.05 moles of the iodide salt. You need to state the salt (KI, NaI, LiI, etc.) in order to determine the actual mass required.
How Wagner react to alkaloid?
Dissolve 2.0 grams of iodine and 6.0 grams of KI in 100.0 ml of H2O.
How do you prepare 40 ml of a 2 mg ml protein solution from a 10mg ml protein solution?
8ml
How would you prepare 385 mL of a 0.766M NaOH solution starting with a 4.03M stock solution?
(385 mL)(0.766 M NaOH) = (X mL)(4.03 M NaOH)4.03X = 294.91X = 73.2 ml============Take 73.2 mL of stock solution and add ( 385 mL - 73.2 mL = 311.8 mL) 312 mL of water to stock solution.
How many mg per ml in a twenty percent solution?
a 2.5% solution has 25mg per ml, therefore a 20% solution has 200mg per ml.
How much 25 mg by ml solution can you make by diluting 120 ml of a 100 mg by ml solution?
480ml
